Concept explainers
Interpretation:
The Lewis structures of the given molecules and ions are to be represented.
Concept Introduction:
In Lewis dot
In Lewis dot symbol, valence electrons are represented by dots.
Dots are placed above and below as well as to the left and right of symbol.
Number of dots is important in Lewis dot symbol but not the order in which the dots are placed around the symbol.
In writing symbol pairing is not done until absolutely necessary.
For metals, the number of dots represents the number of electrons that are lost when the atom forms a cation.
For second period non metals, the number of unpaired dots is the number of bonds the atom can form.
Atomic ions can also be represented by dot symbols, by simply adding (for anions) and subtracting (for cations) the appropriate number of dots from Lewis dot symbol.
Lewis structure is the representation of bonding and non-bonding electron pairs present in the outermost shell of all atoms present in the molecule.
The number of bonds formed by an atom in the molecule is determined by the valence electron pairs.

Answer to Problem 46QP
Solution:
(a)
(b)
(c)
(d)
(e)
(f)
Explanation of Solution
a)
The electronic configuration of nitrogen and chlorine in
The nitrogen atom contains three valence electrons in its
The Lewis structure of
b)
The electronic configuration of oxygen, carbon, and sulfur in
Oxygen and sulfur atoms contain two valence electrons in their
The Lewis structure of
c)
The electronic configuration of oxygen and hydrogen in
The oxygen atom contains four valence electrons in its
The Lewis structure of
d)
The electronic configuration of oxygen, carbon, and hydrogen in
The carbon atom has a tendency to form four bonds because of the presence of four valence electrons in its outermost shell, hydrogen has a tendency to form one bond because of the presence of one electron in its outermost shell, and oxygen has a tendency to form two bonds due to the presence of two electrons in its outermost shell.
The Lewis structure of
e)
The electronic configuration of nitrogen and carbon in
Cyanide ion is composed of one triple bond of carbon and nitrogen atom. This species contains one lone pair on both carbon and nitrogen atoms.
The Lewis structure of
f)
The electronic configuration of carbon, nitrogen, and hydrogen in
Carbon atom has a tendency to form four bonds because of the presence of four valence electrons in its outermost shell, hydrogen has a tendency to form one bond because of the presence of one electron in its outermost shell, and nitrogen has tendency to form four bonds due to the presence of three electrons in its
The Lewis structure of
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Chapter 8 Solutions
BURDGE CHEMISTRY VALUE ED (LL)
- Q5: Label each chiral carbon in the following molecules as R or S. Make sure the stereocenter to which each of your R/S assignments belong is perfectly clear to the grader. (8pts) R OCH 3 CI H S 2pts for each R/S HO R H !!! I OH CI HN CI R Harrow_forwardCalculate the proton and carbon chemical shifts for this structurearrow_forwardA. B. b. Now consider the two bicyclic molecules A. and B. Note that A. is a dianion and B. is a neutral molecule. One of these molecules is a highly reactive compound first characterized in frozen noble gas matrices, that self-reacts rapidly at temperatures above liquid nitrogen temperature. The other compound was isolated at room temperature in the early 1960s, and is a stable ligand used in organometallic chemistry. Which molecule is the more stable molecule, and why?arrow_forward
- A mixture of C7H12O2, C9H9OCl, biphenyl and acetone was put together in a gas chromatography tube. Please decide from the GC resutls which correspond to the peak for C7,C9 and biphenyl and explain the reasoning based on GC results. Eliminate unnecessary peaks from Gas Chromatography results.arrow_forwardIs the molecule chiral, meso, or achiral? CI .CH3 H₂C CIarrow_forwardPLEASE HELP ! URGENT!arrow_forward
- Identify priority of the substituents: CH3arrow_forwardHow many chiral carbons are in the molecule? OH F CI Brarrow_forwardA mixture of three compounds Phen-A, Acet-B and Rin-C was analyzed using TLC with 1:9 ethanol: hexane as the mobile phase. The TLC plate showed three spots of R, 0.1 and 0.2 and 0.3. Which of the three compounds (Phen-A; Acet-B or Rin-C) would have the highest (Blank 1), middle (Blank 2) and lowest (Blank 3) spot respectively? 0 CH: 0 CH, 0 H.C OH H.CN OH Acet-B Rin-C phen-A A A <arrow_forward
- Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning
